We consider the system Applicative_first_order_05__31. Alphabet: !faccolon : [] --> b -> b -> b !facplus : [] --> b -> b -> b a : [] --> b cons : [] --> d -> e -> e false : [] --> c filter : [] --> (d -> c) -> e -> e filter2 : [] --> c -> (d -> c) -> d -> e -> e g : [] --> b -> a -> b map : [] --> (d -> d) -> e -> e nil : [] --> e true : [] --> c Rules: !faccolon (!faccolon x y) z => !faccolon x (!faccolon y z) !faccolon (!facplus x y) z => !facplus (!faccolon x z) (!faccolon y z) !faccolon x (!facplus y (f z)) => !faccolon (g x z) (!facplus y a) map f nil => nil map f (cons x y) => cons (f x) (map f y) filter f nil => nil filter f (cons x y) => filter2 (f x) f x y filter2 true f x y => cons x (filter f y) filter2 false f x y => filter f y Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: Alphabet: !faccolon : [b * b] --> b !facplus : [b * b] --> b a : [] --> b cons : [d * e] --> e false : [] --> c filter : [d -> c * e] --> e filter2 : [c * d -> c * d * e] --> e g : [b] --> a -> b map : [d -> d * e] --> e nil : [] --> e true : [] --> c ~AP1 : [a -> b * a] --> b Rules: !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(~AP1(g(X), Z), !facplus(Y, a)) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) ~AP1(F, X) => F X We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs. After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z)) 1] !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z) 2] !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z) 3] !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z) 4] !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> !faccolon#(~AP1(g(X), Z), !facplus(Y, a)) 5] !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> ~AP1#(g(X), Z) 6] map#(F, cons(X, Y)) =#> F(X) 7] map#(F, cons(X, Y)) =#> map#(F, Y) 8] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 9] filter#(F, cons(X, Y)) =#> F(X) 10] filter2#(true, F, X, Y) =#> filter#(F, Y) 11] filter2#(false, F, X, Y) =#> filter#(F, Y) 12] ~AP1#(F, X) =#> F(X) Rules R_0: !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(~AP1(g(X), Z), !facplus(Y, a)) map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) ~AP1(F, X) => F X Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). The formative rules of (P_0, R_0) are R_1 ::= !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(~AP1(g(X), Z), !facplus(Y, a)) map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) ~AP1(F, X) => F X By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative). Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_1, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) !faccolon#(X, !facplus(Y, ~AP1(F, Z))) >? !faccolon#(~AP1(g(X), Z), !facplus(Y, a)) !faccolon#(X, !facplus(Y, ~AP1(F, Z))) >? ~AP1#(g(X), Z) map#(F, cons(X, Y)) >? F(X) map#(F, cons(X, Y)) >? map#(F, Y) filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) ~AP1#(F, X) >? F(X) !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(~AP1(g(X), Z), !facplus(Y, a)) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) ~AP1(F, X) >= F X We apply [Kop12, Thm. 6.75] and use the following argument functions: pi( ~AP1#(F, X) ) = #argfun-~AP1##(F X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !faccolon = \y0y1.0 !faccolon# = \y0y1.3 + y1 !facplus = \y0y1.2y1 #argfun-~AP1## = \y0.3 + y0 a = 0 cons = \y0y1.1 + y1 + 2y0 false = 3 filter = \G0y1.2y1 filter2 = \y0G1y2y3.1 + 2y2 + 2y3 filter2# = \y0G1y2y3.y3G1(y3) filter# = \G0y1.y1G0(y1) g = \y0y1.0 map = \G0y1.y1 + 2y1G0(y1) map# = \G0y1.3 + y1G0(y1) true = 3 ~AP1 = \G0y1.y1 + G0(y1) ~AP1# = \G0y1.0 Using this interpretation, the requirements translate to: [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 3 + x2 >= 3 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = 3 + x2 >= 3 + x2 = [[!faccolon#(_x1, _x2)]] [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 3 + x2 >= 3 + x2 = [[!faccolon#(_x0, _x2)]] [[!faccolon#(!facplus(_x0, _x1), _x2)]] = 3 + x2 >= 3 + x2 = [[!faccolon#(_x1, _x2)]] [[!faccolon#(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 3 + 2x3 + 2F2(x3) >= 3 = [[!faccolon#(~AP1(g(_x0), _x3), !facplus(_x1, a))]] [[!faccolon#(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 3 + 2x3 + 2F2(x3) >= 3 + x3 = [[#argfun-~AP1##(g(_x0) _x3)]] [[map#(_F0, cons(_x1, _x2))]] = 3 + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) > F0(x1) = [[_F0(_x1)]] [[map#(_F0, cons(_x1, _x2))]] = 3 + F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) >= 3 + x2F0(x2) = [[map#(_F0, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) >= x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = F0(1 + x2 + 2x1) + 2x1F0(1 + x2 + 2x1) + x2F0(1 + x2 + 2x1) >= F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = x2F0(x2) >= x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = x2F0(x2) >= x2F0(x2) = [[filter#(_F0, _x2)]] [[#argfun-~AP1##(_F0 _x1)]] = 3 + max(x1, F0(x1)) > F0(x1) = [[_F0(_x1)]] [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(~AP1(g(_x0), _x3), !facplus(_x1, a))]] [[map(_F0, cons(_x1, _x2))]] = 1 + x2 + 2x1 + 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 1 + x2 + 2x2F0(x2) + 2max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + 2x2 + 4x1 >= 1 + 2x1 + 2x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 1 + 2x1 + 2x2 >= 1 + 2x1 + 2x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 1 + 2x1 + 2x2 >= 2x2 = [[filter(_F0, _x2)]] [[~AP1(_F0, _x1)]] = x1 + F0(x1) >= max(x1, F0(x1)) = [[_F0 _x1]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_1, minimal, formative) by (P_1, R_1, minimal, formative), where P_1 consists of: !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z)) !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z) !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z) !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z) !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> !faccolon#(~AP1(g(X), Z), !facplus(Y, a)) !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> ~AP1#(g(X), Z) map#(F, cons(X, Y)) =#> map#(F, Y) filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) =#> F(X) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_1, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0, 1, 2, 3, 4, 5 * 1 : 0, 1, 2, 3, 4, 5 * 2 : 0, 1, 2, 3, 4, 5 * 3 : 0, 1, 2, 3, 4, 5 * 4 : 0, 1, 2, 3 * 5 : * 6 : 6 * 7 : 9, 10 * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 * 9 : 7, 8 * 10 : 7, 8 This graph has the following strongly connected components: P_2: !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z)) !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z) !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z) !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z) !faccolon#(X, !facplus(Y, ~AP1(F, Z))) =#> !faccolon#(~AP1(g(X), Z), !facplus(Y, a)) P_3: map#(F, cons(X, Y)) =#> map#(F, Y) P_4: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) =#> F(X) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_1, R_1, m, f) by (P_2, R_1, m, f), (P_3, R_1, m, f) and (P_4, R_1, m, f). Thus, the original system is terminating if each of (P_2, R_1, minimal, formative), (P_3, R_1, minimal, formative) and (P_4, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_1, minimal, formative). The formative rules of (P_4, R_1) are R_2 ::= map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(true, F, X, Y) => cons(X, filter(F, Y)) filter2(false, F, X, Y) => filter(F, Y) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_4, R_1, minimal, formative) by (P_4, R_2, minimal, formative). Thus, the original system is terminating if each of (P_2, R_1, minimal, formative), (P_3, R_1, minimal, formative) and (P_4, R_2, minimal, formative) is finite. We consider the dependency pair problem (P_4, R_2, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: filter#(F, cons(X, Y)) >? filter2#(F X, F, X, Y) filter#(F, cons(X, Y)) >? F(X) filter2#(true, F, X, Y) >? filter#(F, Y) filter2#(false, F, X, Y) >? filter#(F, Y) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) filter(F, cons(X, Y)) >= filter2(F X, F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.2 + y0 + y1 false = 3 filter = \G0y1.y1 filter2 = \y0G1y2y3.2 + y2 + y3 filter2# = \y0G1y2y3.1 + y3 + y3G1(y3) filter# = \G0y1.y1 + y1G0(y1) map = \G0y1.y1 + 2y1G0(y1) + 2G0(y1) true = 3 Using this interpretation, the requirements translate to: [[filter#(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > 1 + x2 + x2F0(x2) = [[filter2#(_F0 _x1, _F0, _x1, _x2)]] [[filter#(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2F0(2 + x1 + x2) + x1F0(2 + x1 + x2) + x2F0(2 + x1 + x2) > F0(x1) = [[_F0(_x1)]] [[filter2#(true, _F0, _x1, _x2)]] = 1 + x2 + x2F0(x2) > x2 + x2F0(x2) = [[filter#(_F0, _x2)]] [[filter2#(false, _F0, _x1, _x2)]] = 1 + x2 + x2F0(x2) > x2 + x2F0(x2) = [[filter#(_F0, _x2)]] [[map(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 + 2x1F0(2 + x1 + x2) + 2x2F0(2 + x1 + x2) + 6F0(2 + x1 + x2) >= 2 + x2 + 2x2F0(x2) + 2F0(x2) + max(x1, F0(x1)) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, cons(_x1, _x2))]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 2 + x1 + x2 >= 2 + x1 + x2 = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 2 + x1 + x2 >= x2 = [[filter(_F0, _x2)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_2) by ({}, R_2). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if each of (P_2, R_1, minimal, formative) and (P_3, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_1, minimal, formative). We apply the subterm criterion with the following projection function: nu(map#) = 2 Thus, we can orient the dependency pairs as follows: nu(map#(F, cons(X, Y))) = cons(X, Y) |> Y = nu(map#(F, Y)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_3, R_1, minimal, f) by ({}, R_1, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_1, minimal, formative). The formative rules of (P_2, R_1) are R_3 ::= !faccolon(!faccolon(X, Y), Z) => !faccolon(X, !faccolon(Y, Z)) !faccolon(!facplus(X, Y), Z) => !facplus(!faccolon(X, Z), !faccolon(Y, Z)) !faccolon(X, !facplus(Y, ~AP1(F, Z))) => !faccolon(~AP1(g(X), Z), !facplus(Y, a)) ~AP1(F, X) => F X By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_1, minimal, formative) by (P_2, R_3, minimal, formative). Thus, the original system is terminating if (P_2, R_3, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_3, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(X, !faccolon(Y, Z)) !faccolon#(!faccolon(X, Y), Z) >? !faccolon#(Y, Z) !faccolon#(!facplus(X, Y), Z) >? !faccolon#(X, Z) !faccolon#(!facplus(X, Y), Z) >? !faccolon#(Y, Z) !faccolon#(X, !facplus(Y, ~AP1(F, Z))) >? !faccolon#(~AP1(g(X), Z), !facplus(Y, a)) !faccolon(!faccolon(X, Y), Z) >= !faccolon(X, !faccolon(Y, Z)) !faccolon(!facplus(X, Y), Z) >= !facplus(!faccolon(X, Z), !faccolon(Y, Z)) !faccolon(X, !facplus(Y, ~AP1(F, Z))) >= !faccolon(~AP1(g(X), Z), !facplus(Y, a)) ~AP1(F, X) >= F X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !faccolon = \y0y1.0 !faccolon# = \y0y1.y1 !facplus = \y0y1.y1 a = 0 g = \y0y1.0 ~AP1 = \G0y1.2 + G0(y1) Using this interpretation, the requirements translate to: [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = x2 >= 0 = [[!faccolon#(_x0, !faccolon(_x1, _x2))]] [[!faccolon#(!faccolon(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x1, _x2)]] [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x0, _x2)]] [[!faccolon#(!facplus(_x0, _x1), _x2)]] = x2 >= x2 = [[!faccolon#(_x1, _x2)]] [[!faccolon#(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 2 + F2(x3) > 0 = [[!faccolon#(~AP1(g(_x0), _x3), !facplus(_x1, a))]] [[!faccolon(!faccolon(_x0, _x1), _x2)]] = 0 >= 0 = [[!faccolon(_x0, !faccolon(_x1, _x2))]] [[!faccolon(!facplus(_x0, _x1), _x2)]] = 0 >= 0 = [[!facplus(!faccolon(_x0, _x2), !faccolon(_x1, _x2))]] [[!faccolon(_x0, !facplus(_x1, ~AP1(_F2, _x3)))]] = 0 >= 0 = [[!faccolon(~AP1(g(_x0), _x3), !facplus(_x1, a))]] [[~AP1(_F0, _x1)]] = 2 + F0(x1) >= F0(x1) = [[_F0 _x1]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_3, minimal, formative) by (P_5, R_3, minimal, formative), where P_5 consists of: !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(X, !faccolon(Y, Z)) !faccolon#(!faccolon(X, Y), Z) =#> !faccolon#(Y, Z) !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(X, Z) !faccolon#(!facplus(X, Y), Z) =#> !faccolon#(Y, Z) Thus, the original system is terminating if (P_5, R_3, minimal, formative) is finite. We consider the dependency pair problem (P_5, R_3, minimal, formative). We apply the subterm criterion with the following projection function: nu(!faccolon#) = 1 Thus, we can orient the dependency pairs as follows: nu(!faccolon#(!faccolon(X, Y), Z)) = !faccolon(X, Y) |> X = nu(!faccolon#(X, !faccolon(Y, Z))) nu(!faccolon#(!faccolon(X, Y), Z)) = !faccolon(X, Y) |> Y = nu(!faccolon#(Y, Z)) nu(!faccolon#(!facplus(X, Y), Z)) = !facplus(X, Y) |> X = nu(!faccolon#(X, Z)) nu(!faccolon#(!facplus(X, Y), Z)) = !facplus(X, Y) |> Y = nu(!faccolon#(Y, Z)) By [FuhKop19, Thm. 61], we may replace a dependency pair problem (P_5, R_3, minimal, f) by ({}, R_3, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++ [FuhKop19] C. Fuhs, and C. Kop. A static higher-order dependency pair framework. In Proceedings of ESOP 2019, 2019. [Kop11] C. Kop. Simplifying Algebraic Functional Systems. In Proceedings of CAI 2011, volume 6742 of LNCS. 201--215, Springer, 2011. [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.